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Let $X_1,X_2 $ and $X_3$ be a random sample from $N(3,12)$ distribution. If $\bar X=\dfrac{1}{3} \sum_{i=1}^{3}X_i$ and $S^2=\dfrac{1}{2}\sum_{i=1}^{3}(X_i-\bar X)^2$ denote the sample mean and the sample variance respectively, then $P(1.65<\bar X \leq4.35 ,.12<S^2 \leq55.26) $ is

$(A)=.49$

$(B)=.50$

$(C)=.98$

$(D)={}$None of the above

My first question is what does **

$P(1.65<\bar X \leq4.35 ,.12< S^2\leq55.26)$

** mean? Is it $P(1.65<\bar X \leq4.35 ,.12<S^2 \leq55.26)=P(1.65<\bar X \leq4.35)\cap P(.12<S^2 \leq55.26)$?

I have doubt in this very first step also i have done some work after that assuming(the step above mentioned) and i want to ask about that too but that i will ask if this step is fine. I will edit my question but before that please tell me about it. (My English is weak so please ignore grammar mistakes).

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Hints:

As the sample are independent identically normally distributed r.v.s, thus $S^2 = \frac{1}{n-1}\sum (X_i - \bar{X})^2$ ans $\bar{X}_n$ are independent as well (this is not a trivial claim, but I guess in your context you can use it as a fact) and recall that $$ \bar{X}_n \sim \mathcal{N}(\mu, \sigma^2/n), \quad \frac{S^2}{\sigma^2} \sim \frac{1}{n-1}\chi^2_{(n-1)}. $$ And recall that if $A$ and $B$ are independent events then $$ \mathbb{P}(A\cap B)= \mathbb{P}(A)\mathbb{P}(B). $$

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  • $\begingroup$ Yes but that step is next to my doubt . How to deal with that thing i mentioned in my question ? $\endgroup$ – Daman Jan 29 '18 at 11:27
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    $\begingroup$ Well, if the r.v.s are independent then $P(1.65<\bar X \leq4.35 ,12<S^2 \leq55.26)=P(1.65<\bar X \leq4.35) P(12<S^2 \leq55.26)$ $\endgroup$ – V. Vancak Jan 29 '18 at 12:28
  • $\begingroup$ Thank you very much i had doubt in that portion. I am more familiar with $\cap$ between two variables. $\endgroup$ – Daman Jan 29 '18 at 12:50

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